What is the value of $\dfrac{d}{dx}(-2x^4+3x^3-x^2)$ at $x=-2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $104$ (Choice B) B $30$ (Choice C) C $-16$ (Choice D) D $-60$
Explanation: Let's first find the expression for $\dfrac{d}{dx}(-2x^4+3x^3-x^2)$ and then evaluate it at $x=-2$. According to the sum rule, the derivative of $-2x^4+3x^3-x^2$ is the sum of the derivatives of $-2x^4$, $3x^3$, and $-x^2$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(-2x^4)&=-2\dfrac{d}{dx}(x^4)&&\gray{\text{Constant multiple rule}}\\\\ &=-2\cdot (4x^3)&&\gray{\text{Power rule}}\\ \\ &=-8x^3\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(-2x^4+3x^3-x^2) \\\\ &=-2\dfrac{d}{dx}(x^4)+3\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(x^2)&&\gray{\text{Basic differentiation rules}} \\\\ &=-2\cdot 4x^3+3\cdot3x^2-2x&&\gray{\text{The power rule}} \\\\ &=-8x^3+9x^2-2x \end{aligned}$ So we found that $\dfrac{d}{dx}(-2x^4+3x^3-x^2)=-8x^3+9x^2-2x$. Plugging in $x=-2$ and evaluating using the calculator, we find that the value is $104$. In conclusion, the value of $\dfrac{d}{dx}(-2x^4+3x^3-x^2)$ at $x=-2$ is $104$.